The limit of [math]f\left(x\right)[/math] as [math]x\to c[/math] is [math]L[/math] if and only if for every distance, [math]\epsilon>0[/math], there is another distance, [math]\delta>0[/math], such that if [i]x[/i] is within delta of [i]c[/i], [math]\left|x-c\right|<\delta[/math], then [i]f[/i]([i]x[/i]) is within epsilon of [i]f[/i]([i]c[/i]), [math]\left|f\left(x\right)-f\left(c\right)\right|<\epsilon[/math].[br][br]This game will help you understand the statements above.

The function below is [math]f\left(x\right)=\left(x^2-16\right)\left(x+2\right)+43[/math][br]We claim that [math]\lim_{_{x\to-2}}f\left(x\right)=43[/math]. [br]The epsilon is given. Try to find a suitable delta (small enough), so that the image of the set of all x's such that [math]\left|x--2\right|<\delta[/math] (which are represented by where the graph of [i]f[/i] intersects the green vertical band), are within the set of [i]f[/i]([i]x[/i])'s such that [math]\left|f\left(x\right)-43\right|<\epsilon[/math] (which is represented by where the graph of [i]f[/i] intersects the red horizontal band)

The function below is [math]f\left(x\right)=\left(x^2-16\right)\left(x+2\right)+43[/math][br]We claim that [math]\lim_{_{x\to-2}}f\left(x\right)=43[/math]. [br]The epsilon is given. Try to find a suitable delta (small enough), so that the image of the set of all x's such that [math]\left|x--2\right|<\delta[/math] (which are represented by where the graph of [i]f[/i] intersects the green vertical band), are within the set of [i]f[/i]([i]x[/i])'s such that [math]\left|f\left(x\right)-43\right|<\epsilon[/math] (which is represented by where the graph of [i]f[/i] intersects the red horizontal band)